A proof using $\Delta^ny(t)=\sum_{k=0}^{n}{(-1)^k\binom{n}{k}y(t+n-k)}$ Please How can I use $\Delta^ny(t)=\sum_{k=0}^{n}{(-1)^k\dbinom{n}{k}y(t+n-k)}$ to prove
$\sum_{i=0}^{n}{(-1)^i\dbinom{n}{i}y(i)}=(-1)^n\Delta^ny(0)$ and hence to evaluate
$\sum_{i=0}^{n}{\dfrac{(-1)^i}{2+i}\dbinom{n}{i}}$?
For the evaluation of $\sum_{i=0}^{n}{\dfrac{(-1)^i}{2+i}\dbinom{n}{i}}$, I know all I have to do is to take $y(i)=\dfrac{1}{2+i}$ and find $\Delta^n\left(\dfrac{1}{2+i}\right)$ for $i=0$. So how can I find this $\Delta^n\left(\dfrac{1}{2+i}\right)$.
Thanks
 A: For the first part let $k=n-i$:
$$\sum_{i=0}^n(-1)^i\binom{n}iy(i)=\sum_{i=0}^n(-1)^i\binom{n}{n-i}y(i)=\sum_{k=0}^n(-1)^{n-k}\binom{n}ky(-k)\;.$$
Now note that $(-1)^{n-k}=(-1)^{n+k}$, and you have
$$\sum_{i=0}^n(-1)^i\binom{n}iy(i)=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}ky(0+n-k)=(-1)^n\Delta^ny(0)\;.$$
Improved version: For the second part, note that
$$\Delta^n\left(\frac1{x+1}\right)=\Delta^nx^{\underline{-1}}=(-1)^nn!x^{\underline{-(n+1)}}=\frac{(-1)^nn!}{(x+1)^{\overline{n+1}}}\;.$$
Substitute $x=1$ to see that
$$\Delta^ny(0)=\frac{(-1)^nn!}{2^{\overline{n+1}}}=\frac{(-1)^nn!}{(n+2)!}=\frac{(-1)^n}{(n+1)(n+2)}\;.$$
Original version: I expect that there’s an easier way to do the second part, but I’m too tired right now to look for it; here’s a way that works. A little computation suggests the conjecture that
$$(-1)^n\Delta^ny(0)=\frac1{(n+1)(n+2)}\;.$$
Now
$$\begin{align*}
(-1)^n\Delta^ny(0)&=\sum_{k=0}^n\frac{(-1)^k}{k+2}\binom{n}k\\\\
&=\sum_{k=0}^n\frac{(-1)^k(k+1)}{(n+1)(n+2)}\binom{n+2}{k+2}\\\\
&=\frac1{(n+1)(n+2)}\sum_{k=0}^n(-1)^k(k+1)\binom{n+2}{k+2}\;,
\end{align*}$$
so the conjecture is proved if we can show that
$$\sum_{k=0}^n(-1)^k(k+1)\binom{n+2}{k+2}=1\;.\tag{1}$$
Now
$$\begin{align*}
\sum_{k=0}^n(-1)^k(k+1)\binom{n+2}{k+2}&=\sum_{k=0}^n(-1)^k\Big((k+2)-1\Big)\binom{n+2}{k+2}\\\\
&=\sum_{k=0}^n(-1)^k(k+2)\binom{n+2}{k+2}-\sum_{k=0}^n(-1)^k\binom{n+2}{k+2}\\\\
&=(n+2)\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}-\sum_{k=2}^{n+2}(-1)^k\binom{n+2}k\;,
\end{align*}$$
by the distributive law.
and you can evaluate each of the summations in $(1)$ by using the fact that
$$\sum_{k=0}^m(-1)^k\binom{m}k=0$$
for $m>0$; if you do it correctly, you’ll establish $(1)$.
A: Just note that, by the binomial definition, we have
$$\Delta^ny(t)= \sum_{k=0}^{n}{(-1)^k\dbinom{n}{k}y(t+n-k)}$$ 

$$ =\sum_{k=0}^{n}{(-1)^{n-k}\dbinom{n}{k}y(t+k)}=(-1)^n \Delta^n y(t). $$

Use the highlighted equation to answer the first part by substituting $t=0$. 
Added: For the second part, you can use the technique as 

$$ \sum_{i=0}^{n}(-1)^i{ n \choose i}x^{i+1}=x(1-x)^n. $$

Now, integrating both sides w.r.t. $x$ from $0$ to $1$ gives 

$$ \sum_{i=0}^{n}(-1)^i{ n \choose i}\frac{1}{i+2} = \int_{0}^{1}x(1-x)^n\, dx =\frac{1}{(n+1)(n+2)}. $$

The above integral can be evaluated using integration by with parts with $u=x$.
Note: 

$$ (x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}.  $$

